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Calculate the standard enthalpy change for the vaporization of 15 grams of water given that the molar enthalpy of vaporization of water is 40.7 kJ/mol.

asked 4 days ago in Chemical thermodynamics by AllieNoyes3 (410 points)

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Calculate the standard enthalpy change for the vaporization of 15 grams of water at its boiling point, given that the molar enthalpy of vaporization of water is 40.7 kJ/mol.

asked 4 days ago in Chemical thermodynamics by AstridCrooke (410 points)

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Calculate the standard enthalpy change for the vaporization of 100 g of water at 100°C given that the enthalpy of vaporization of water is 40.7 kJ/mol.

asked 4 days ago in Chemical thermodynamics by KateNewcombe (510 points)

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Calculate the standard enthalpy change for the vaporization of 10 grams of water at 100°C, given that the enthalpy of vaporization for water is 40.7 kJ/mol.

asked 4 days ago in Chemical thermodynamics by IngridMarcha (430 points)

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Calculate the standard enthalpy change for the vaporization of 1 mole of water at 100°C, given that the enthalpy of fusion of water is 6.01 kJ/mol and the enthalpy of vaporization of water is 40.7 kJ/mol.

asked 4 days ago in Chemical thermodynamics by ValeriaMetca (450 points)

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Calculate the standard enthalpy change for the vaporization of 1 mole of water at 100°C and 1 atm pressure. The enthalpy of formation of liquid water is -285.8 kJ/mol and the enthalpy of formation of water vapor is -241.8 kJ/mol.

asked 4 days ago in Chemical thermodynamics by SheriFkw4940 (630 points)

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Calculate the standard enthalpy change for the transition of liquid water at 100°C to steam at the same temperature, given that the enthalpy of vaporization of water is 40.7 kJ mol-1.

asked 4 days ago in Chemical thermodynamics by AurelioMetts (450 points)

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Calculate the standard enthalpy change for the sublimation of iodine, given that the sublimation of 1 mole of iodine requires 62.44 kJ of energy and the standard enthalpy of formation of iodine (s) is 0 kJ/mol.

asked 4 days ago in Chemical thermodynamics by Kristian7404 (470 points)

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Calculate the standard enthalpy change for the sublimation of iodine solid to iodine gas at 298 K with the following given data: - Standard enthalpy of fusion of iodine: 15.52 kJ/mol- Standard molar entropy of iodine solid: 62.7 J/K/mol- Standard molar entropy of iodine gas: 260.6 J/K/mol

asked 4 days ago in Chemical thermodynamics by Shirley80008 (530 points)

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Calculate the standard enthalpy change for the sublimation of iodine at 25°C, given that the standard enthalpy of fusion of iodine is 15.7 kJ/mol and the standard enthalpy of vaporization of iodine is 41.0 kJ/mol. (The molar mass of iodine is 126.90 g/mol)

asked 4 days ago in Chemical thermodynamics by ErnaGrimshaw (490 points)

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Calculate the standard enthalpy change for the sublimation of iodine (I2) using the given information: The standard enthalpy of fusion of iodine is 15.52 kJ/mol and the standard enthalpy of vaporization of iodine is 41.57 kJ/mol.

asked 4 days ago in Chemical thermodynamics by BartLundie14 (570 points)

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Calculate the standard enthalpy change for the sublimation of iodine (I2) if 1 mole of I2(s) is converted to 1 mole of I2(g) at 25°C and 1 bar pressure. Given the following information:- The standard enthalpy of fusion of I2 is 15.4 kJ/mol- The standard enthalpy of vaporization of I2 is 41.3 kJ/mol- The standard entropy change for the sublimation of I2 is 62.4 J/mol K.

asked 4 days ago in Chemical thermodynamics by LindsayI610 (390 points)

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Calculate the standard enthalpy change for the sublimation of iodine (I2) given that the enthalpy of fusion for iodine is 15.52 kJ/mol and the enthalpy of vaporization for iodine is 41.57 kJ/mol.

asked 4 days ago in Chemical thermodynamics by IngridWills (350 points)

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Calculate the standard enthalpy change for the sublimation of 5 moles of solid iodine (I2) given the following data:- Enthalpy of fusion of iodine = 15.52 kJ/mol- Enthalpy of vaporization of iodine = 41.57 kJ/mol

asked 4 days ago in Chemical thermodynamics by ShelliFogart (370 points)

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Calculate the standard enthalpy change for the reduction of iron(III) oxide to iron using the following balanced chemical equation:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)Given that the standard enthalpy change of formation for Fe2O3(s) is -824.2 kJ/mol, the standard enthalpy change of formation for CO2(g) is -393.5 kJ/mol, and the standard enthalpy change of formation for Fe(s) is 0 kJ/mol. Assume all reactants and products are in their standard states.

asked 4 days ago in ThermoChemistry by KoryLongstaf (330 points)

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Calculate the standard enthalpy change for the reduction of iron (III) oxide using carbon monoxide as the reducing agent, given that the standard enthalpy of formation for iron (III) oxide is -824.2 kJ/mol and the standard enthalpy of formation for carbon monoxide is -110.5 kJ/mol.

asked 4 days ago in ThermoChemistry by PWOMazie102 (650 points)

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Calculate the standard enthalpy change for the reaction:Cu(H2O)62+ (aq) + 4Cl-(aq) → CuCl42- (aq) + 12H2O (l)given the following standard enthalpies of formation:Cu(H2O)62+ (aq): -1846.4 kJ/molCuCl42- (aq): -3599.5 kJ/molH2O (l): -285.8 kJ/molCl- (aq): -167.2 kJ/mol

asked 4 days ago in Chemical thermodynamics by InaLewers025 (330 points)

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Calculate the standard enthalpy change for the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) given the following enthalpy changes: ∆Hf° for NaCl(aq) = -407.3 kJ/mol ∆Hf° for H2O(l) = -285.8 kJ/mol ∆Hf° for NaOH(aq) = -469.11 kJ/mol ∆Hf° for HCl(aq) = -167.16 kJ/mol

asked 4 days ago in Chemical thermodynamics by MagdaLanglai (610 points)

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Calculate the standard enthalpy change for the reaction: 2Fe(s) + 3/2O2(g) → Fe2O3(s), given that the standard enthalpy of formation for Fe2O3(s) is -824.2 kJ/mol, the standard enthalpy of formation for Fe(s) is 0 kJ/mol, and the standard enthalpy of formation for O2(g) is 0 kJ/mol.

asked 4 days ago in Chemical thermodynamics by RWTIona27307 (670 points)

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Calculate the standard enthalpy change for the reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Given the following standard enthalpy of formation values: ΔHf° [Fe2O3(s)] = -824 kJ/mol ΔHf° [CO(g)] = -110 kJ/mol ΔHf° [CO2(g)] = -393.5 kJ/mol ΔHf° [Fe(s)] = 0 kJ/mol.

asked 4 days ago in Chemical thermodynamics by EldenNobbs11 (450 points)

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